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Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1 / 2 2 / \ / 3 4 4 3
But the following is not:
1 / 2 2 \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
对称的递归表达式：
testMirror(leftNode->left,rightNode->right)&&testMirror(leftNode->right,rightNode->left);
leftNode->left->val==rightNode->right->val
leftNode->right->val==rightNode->left->val
1 class Solution { 2 3 public: 4 5 bool isSymmetric(TreeNode *root) { 6 7 if(root==NULL) 8 9 { 10 11 return true; 12 13 } 14 15 return testMirror(root->left,root->right); 16 17 18 19 } 20 21 22 23 bool testMirror(TreeNode *leftNode,TreeNode *rightNode) 24 25 { 26 27 28 29 if(leftNode==NULL&&rightNode==NULL) 30 31 return true; 32 33 if(leftNode!=NULL&&rightNode==NULL||leftNode==NULL&&rightNode!=NULL) 34 35 return false; 36 37 if(leftNode->val!=rightNode->val) 38 39 return false; 40 41 42 43 return testMirror(leftNode->left,rightNode->right)&&testMirror(leftNode->right,rightNode->left); 44 45 46 47 48 49 } 50 51 };
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原文地址：http://www.cnblogs.com/reachteam/p/4251662.html
